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ANALISIS STRUKTUR PORTAL BERGOYANG
Oleh :Ardia Tiara Rahmi
Muhammad laul HudaRatna Nastiti NugraeniShosy Barutama Putra
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Determine each reaction s for the frame illustrated below using moment distribution . Then draw the shear force and bending mmoment diagrams for the frame.
QUESTION
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Nilai Kekakuan
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Titik E
Titik D Titik F
Koefisien Distribusi
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Mo (Akibat beban)
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M’ (Akibat goyangan)
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Tabel Perataan Momen Akibat Beban
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Tabel Perataan Momen Akibat BebanTabel Perataan Momen Akibat Goyangan
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Mencari nilai ɱAkibat Beban
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Akibat Goyangan
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Sehingga momen akibat goyangan menjadi…
Dan momen total menjadi…
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FREE BODY
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Bidang M (Momen)Bidang MMx dari D - E
3
qq’
x’
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Bidang M (Momen) Batang BE Mb = - 0,31 tm Me = + 0,44 tm Batang EF Me = - 0,98 M1 = 1,04x2-0,98
= 1,10 tm Mf = - 1,87 tm Batang F2 Mf = -2 tm M2 = 0 Batang FC Mf = 0,13 tm Mc = 0,01 tm
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Bidang M (Momen)
0,16
1,87
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Bidang D (Gaya Lintang)
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Batang BE• Db1 = 0• Db2 = 0,25 t =De1• De2 = 0Batang EF• De1 = 0• De2 = 1,04 t =D11• D12 =1,04-4 = -2,96 t
= Df1• Df2 =0
Batang F2• Df1 =0• Df2 = 2t• D2 =0Batang FC• Df1 =0• Df2 = -0,02 =Dc1• Dc2 = 0
Bidang D (Gaya Lintang)
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Bidang D (Gaya Lintang)
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Batang ADNa1 =0Na2 =1,2t =Nd1Nd2 =0
Batang DENd1 =0Nd2 =0,23t =Ne1Ne2 =0
Batang BENb1 =0Nb2 =4,34t =Ne1Ne2 =0
Bidang N (Gaya Normal)Batang EFNe1 =0Ne2 =0,02t =Nf1Nf2 =0
Batang F2Nf1 =0Nf2 =0 =N21N22 =0
Batang FCNf1 =0Nf2 =4,96t =Nc1Nc2 =0
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Bidang N (Gaya Normal)
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Thank You
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